Matrices :D :D :D
by utkarshgupta, Jul 4, 2016, 3:56 AM
Long time people 
This sure is nice and I wouldn't have been able to solve it without first solving the
version.
Problem (Romanian Mathematical Olympiad Grade XI) :
Let
be a real 
matrix, such that 
, for all 
. Prove that for all non-negative real numbers 
we have ![\[ \det(A+xI_n)\cdot \det(A+yI_n) \geq \det (A+\sqrt{xy}I_n)^2.\]](//latex.artofproblemsolving.com/f/c/c/fcc03110bf8f741c4be0a64df336378ca84802e4.png)
Solution :
Let
Obviously
is a polynomial.
So we have to show that
But the above result will follow directly once we establish that all the coefficients of are positive.
obviously.
I will call the kind of matrices asked in the question as ski matrix.
To show this I will use induction on the order of the matrix.
Let the result be true for all such
ski matrices.
Now consider any
such ski matrix and call it 
.
Then,
But obviously since 
is a sum of determinants of ski matrices of the order , all it's coefficients by the induction hypothesis are positive and hence so are it's integrals (except the constants which we have already established as zero).
Hence
has all it's coefficients positive and hence by Cauchy Schwartz inequality, we are done.
This sure is nice and I wouldn't have been able to solve it without first solving the
version.Problem (Romanian Mathematical Olympiad Grade XI) :
Let
be a real 
matrix, such that 
, for all 
. Prove that for all non-negative real numbers 
we have ![\[ \det(A+xI_n)\cdot \det(A+yI_n) \geq \det (A+\sqrt{xy}I_n)^2.\]](http://latex.artofproblemsolving.com/f/c/c/fcc03110bf8f741c4be0a64df336378ca84802e4.png)
Solution :
Let
Obviously
is a polynomial.So we have to show that
But the above result will follow directly once we establish that all the coefficients of
are positive.
obviously.I will call the kind of matrices asked in the question as ski matrix.
To show this I will use induction on the order of the matrix.
Let the result be true for all such
ski matrices.Now consider any
such ski matrix and call it 
.Then,
But obviously since 
is a sum of determinants of ski matrices of the order , all it's coefficients by the induction hypothesis are positive and hence so are it's integrals (except the constants which we have already established as zero).Hence
has all it's coefficients positive and hence by Cauchy Schwartz inequality, we are done.
September 2017
February 2017